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  • If ΔABC is right angled at C, then the value of cos (A+B) is (A) 0 (B) 1 (C) 1 by 2 (D) sqroot 3 by 2

If ΔABC is right angled at C, then the value of cos (A+B) is

(A) 0 \ \ \ \ \ (B) 1 \ \ \ \ \ (C)1/2 \ \ \ \ \ (D)\frac{\sqrt{3}}{2}                 

Answers (1)

Answer.     [A]                  
Solution.     It is given that \angleC = 90°
               

In \bigtriangleupABC

\angleA +\angleB +\angleC = 180             [\because  sum of interior angles of triangle is 180°]
\angleA + \angleB + 90o = 180            [\because C = 90° (given)]
\angleA + \angleB = 1800 – 900
\angleA + \angleB = 90°          …(1)
cos(\angleA +\angleB) = cos (90°)
cos(90°) = 0
[\because from the table of trigonometric ratios of angles we know that cos 90° = 0]
Hence option A is correct.

 

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